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## #51 2013-03-05 06:09:41

anonimnystefy
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### Re: Kummer and a tough integral.

Yes, it does.

But, what does it mean?

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #52 2013-03-05 06:14:20

bobbym

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### Re: Kummer and a tough integral.

Hmmm? Change that exp(-x) to sqrt(x) and change f(0,50 ) to f(0,1)

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #53 2013-03-05 06:19:48

anonimnystefy
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### Re: Kummer and a tough integral.

It says "Romberg failed to converge".

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #54 2013-03-05 06:21:27

bobbym

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### Re: Kummer and a tough integral.

Try adjusting rombergtol: 1.e-3

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #55 2013-03-05 06:22:55

anonimnystefy
Real Member

Offline

### Re: Kummer and a tough integral.

It is working and giving out: 0.66653268137515

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #56 2013-03-05 06:28:03

bobbym

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### Re: Kummer and a tough integral.

There should be a global variable called rombergit. What is its value?

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #57 2013-03-05 06:33:06

anonimnystefy
Real Member

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### Re: Kummer and a tough integral.

It is 11.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #58 2013-03-05 06:35:18

bobbym

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### Re: Kummer and a tough integral.

Set it  25. and rombergtol: 1.e-6, what happens?

You should know about these:

http://eagle.cs.kent.edu/MAXIMA/maxima_21.html

http://www.csulb.edu/~woollett/

get them if you do not.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #59 2013-03-05 07:03:18

anonimnystefy
Real Member

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### Re: Kummer and a tough integral.

I get 0.66666657420034.

Thanks for the links. I will try 'em out.

Last edited by anonimnystefy (2013-03-05 07:06:28)

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #60 2013-03-05 07:08:59

bobbym

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### Re: Kummer and a tough integral.

Here is more:

http://maxima.sourceforge.net/documentation.html

I got
romberg(1/(cos(x)+x^2), x, 0, 100);
1.828017700397752

which is accurate. By setting rombergit to 200 and rombergtol to 1E-10.

Now you need to get a bfloat in there somehow.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #61 2013-04-03 06:41:27

anonimnystefy
Real Member

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### Re: Kummer and a tough integral.

Hi bobbym

How do I get more accuracy out of NIntegrate?

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #62 2013-04-03 06:43:06

bobbym

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### Re: Kummer and a tough integral.

Easy, use WorkingPrecision-> some number.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #63 2013-04-03 06:51:05

anonimnystefy
Real Member

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### Re: Kummer and a tough integral.

Ah, there it is. I found the AccuracyGoal and PrecisionGoal, but they didn't help.

Thanks!

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #64 2013-04-03 06:55:58

bobbym

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### Re: Kummer and a tough integral.

WorkingPrecision did not work?

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #65 2013-04-03 07:00:23

anonimnystefy
Real Member

Offline

### Re: Kummer and a tough integral.

It is working.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #66 2013-04-03 07:00:48

bobbym

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### Re: Kummer and a tough integral.

The answer to d) i) is in the other thread.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.